By definition, an infinite product converges if its sequence of finite partial products converges to a nonzero number. Find all \(z\in\mathbb C\) for which \[ \prod_{k=1}^{\infty}(1-z^k) \] converges.
Solution
For \(n\ge1\), write \[ P_n(z)\coloneqq\prod_{k=1}^n(1-z^k). \]
1. If the infinite product converges in the stated sense, then\[ z^n\longrightarrow0. \]
Suppose \[ P_n(z)\longrightarrow P \] with \(P\ne0\). Then also \[ P_{n-1}(z)\longrightarrow P. \] For all sufficiently large \(n\), \(P_{n-1}(z)\ne0\), and therefore \[ 1-z^n = \frac{P_n(z)}{P_{n-1}(z)} \longrightarrow \frac PP = 1. \] Hence \(z^n\to0\).
2. The conclusion of step 1 implies\[ |z|<1. \]
If \(|z|\ge1\), then \[ |z^n|=|z|^n\ge1 \] for every \(n\), so \(z^n\) cannot tend to \(0\). Thus convergence of the product requires \(|z|<1\).
3. Now suppose \(|z|<1\). There is \(N\) such that for every \(k\ge N\),\[ |z|^k\le\frac12 \]and\[ \left|\log(1-z^k)\right| \le 2|z|^k, \]where \(\log\) denotes the principal logarithm.
Since \(|z|^k\to0\), choose \(N\) so that \(|z|^k\le1/2\) for \(k\ge N\). For \(|w|<1\), \[ \log(1-w) = -\sum_{m=1}^{\infty}\frac{w^m}{m}. \] Consequently, for \(|w|\le1/2\), \[ \begin{aligned} |\log(1-w)| &\le \sum_{m=1}^{\infty}\frac{|w|^m}{m}\\ &\le \sum_{m=1}^{\infty}|w|^m\\ &= \frac{|w|}{1-|w|}\\ &\le 2|w|. \end{aligned} \] Apply this with \(w=z^k\).
4. If \(|z|<1\), then\[ \sum_{k=N}^{\infty}\log(1-z^k) \]converges absolutely.
By step 3, \[ \sum_{k=N}^{\infty} |\log(1-z^k)| \le 2\sum_{k=N}^{\infty}|z|^k. \] The series on the right is geometric and converges because \(|z|<1\).
5. If \(|z|<1\), the partial products \(P_n(z)\) converge to a nonzerolimit.
4. For \(n\ge N\),\[ \prod_{k=N}^n(1-z^k) = \exp\left( \sum_{k=N}^n\log(1-z^k) \right). \]Taking \(n\to\infty\) gives\[ \prod_{k=N}^{\infty}(1-z^k) = e^L \ne 0. \]The finite initial product\[ \prod_{k=1}^{N-1}(1-z^k) \]is also nonzero, so multiplying it by \(e^L\) gives a nonzero limit for\(P_n(z)\).
6. Therefore the product converges exactly for\[ \boxed{\{z\in\CC:|z|<1\}}. \]
Steps <1>1–<1>2 prove necessity, and steps <1>3–<1>5 prove sufficiency.
7. Q.E.D.
Step 6 gives the requested set.